Question

A gas occupies 400.0 mL at 20.00 C and 70.00 kPa. What will the pressure be at 40.00 C and 200.0 mL ?

Answer 1

Answer: The new pressure is 280 kPa.

Step-by-step explanation:

Given:
`V_(1)` = 400.0 mL,
`T_(1) = 20^(o)C` ,
`P_(1)` = 70.0 kPa

`V_(2)` = 200.0 mL,
`T_(2) = 40.0^(o)C`,
`P_(2)` = ?

Now, combined gas law is used to calculate the new pressure as follows.

`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`

Substitute the values into above formula as follows.

`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))\\(70.0 kPa * 400.0 mL)/(20^(o)C) = (P_(2) * 200.0 mL)/(40.0^(o)C)\\P_(2) = 280 kPa`

Thus, we can conclude that the new pressure is 280 kPa.